How to get the explicit result of this limit?

Question

Let $f(x) = x^2+1$. I would like to compute the limit $$ \lim_{k \to +\infty} k \int_0^1 \sin \left [ \left ( k + \frac{1}{2} \right ) \pi x \right ] f(x) \, \mathrm d x. $$

Then I use the command

Limit[Integrate[k * Sin[( k + 1 / 2) * Pi * x] * (x^2 + 1), {x, 0, 1}], k -> Infinity]

Mathematica returns Indeterminate

Could you explain how to get the explicit limit? Thank you so much for your help!

Answer 1

DiscreteLimit[k Integrate[Sin[(k + 1/2)*Pi*x]*(x^2 + 1), {x, 0, 1}], 
 k -> Infinity]

1/π

Also,

Series[Integrate[k Sin[(k + 1/2)*Pi*x]*(x^2 + 1), {x, 0, 1}], {k, 
   Infinity, 0}] // Normal

gives almost the answer

1/π + (2 Sin[k π])/π

Edit: From the documentation we read

DiscreteLimit is also known as discrete limit or limit over the integers.

which is my understanding of the basic difference between this and Limit

Edit 2: Thanks to @BobHanlon

The corresponding Series to the DiscreteLimit is given by

Series[Integrate[k Sin[(k + 1/2)*Pi*x]*(x^2 + 1), {x, 0, 1}], {k, Infinity, 0}, Assumptions -> k ∈ PositiveIntegers] // Normal

and evaluates to the expected result.

Answer 2

No need to only consider Element[k,Integers]

Integrate[k Sin[(k + 1/2)*Pi*x]*(x^2 + 1), {x, 0, 1}] 
Asymptotic[%, k -> Infinity]// Collect[#, k] &
  

$\frac{2 \sin (\pi k)}{\pi }+\frac{-\frac{\sin (\pi k)}{\pi }+\frac{2 \cos(\pi k)}{\pi ^2}-\frac{1}{2 \pi}}{k}+\frac{1}{\pi }$